Problem: $f(x, y) = 4x - y + \cos(2x)$ What are all the critical points of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(\pi, 4\pi + 1)$ (Choice B) B $\left( 4 + \dfrac{\cos(2)}{2}, -1 \right)$ (Choice C) C $(0, 0)$ (Choice D) D There are no critical points.
Answer: A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} 4 - 2\sin(2x) \\ -1 \end{bmatrix}$ Notice that the $y$ -component of the gradient is always $-1$. In other words, no matter what input we feed the gradient of $f$, it will never equal the zero vector. Therefore, there are no critical points.